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If V={(x,y)|x,y ∈ R} with acts (x1,y1)@(x2,y2)=\((\sqrt[3]{(x1^3+x2^3)},\sqrt[3]{(y1^3+y2^3)})\) and a$(x1,y1)=(ax1,ay1) , a ∈ R.

Prove V with this acts is NOT vector space.Thank you! 

 Nov 12, 2018
 #1
avatar+6046 
+1

are you sure it's not?

 

it's addition has commutativity and associativity

 

(0,0) is the clear additive identity and (-x,-y) is the additive inverse of (x,y)

 

the scalar properties are identical to the Euclidean vector space

 

I'm not seeing why this isn't a vector space.

 

remember the cube and cube root form a bijection

 Nov 12, 2018
 #2
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This is not a vector space. Exactly one of the axioms of a vector space is not satisfied.

Guest Nov 12, 2018
 #3
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Rom yes im sure the exercise say it's not.For this reason i can't solve the exercise

Dimitristhym  Nov 12, 2018
 #4
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Guest "Exactly one of the axioms of a vector space is not satisfied"

Witch is this axiom? 

Dimitristhym  Nov 12, 2018
 #5
avatar+6046 
+3

Ok.  Distribution of scalar mutliplication with respect to field addition doesn't hold.

 

we should have

 

\((a+b)\vec{v} = a \vec{v} \oplus b \vec{v}\)

 

let's take a look

 

\((a+b)\vec{v} =\{(a+b)v_x,(a+b)v_y\}\\ a\vec{v} \oplus b \vec{v} =\left \{\sqrt[3]{a^3 v_x^3+b^3 v_x^3}, \sqrt[3]{a^3 v_y^3+b^3 v_y^3 }\right \}=\\ \left\{\sqrt[3]{a^3+b^3}v_x,~\sqrt[3]{a^3+b^3}v_y \right\}\\ \sqrt[3]{a^3+b^3} \neq a+b\)

Rom  Nov 13, 2018
 #6
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Thank you very much Rom! 

Dimitristhym  Nov 13, 2018

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